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At what speeds will the steady state amp of torsional oscillations of the disc of the system be less than 2 degree?
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G = $80 \ \times \ 10^9 \ N/m^2$

J = $1.8 \ \times \ 10^{-6} \ m^4$

$\theta \ = 2˚ = 2 \times \frac{\pi }{180} \ = \ 0.0349$ rad.

$ - kt \ = \ \frac{GJ}{L} = \frac{80 \ \times \ 10^9 \ \times \ 1.8 \times \ 10^{-6}}{0.6}$

$kt \ = \ 240 \ \times \ 10^3$ Nm/rad

$W_n \ = \ \sqrt{ \frac{Kt}{I}}$

$= \sqrt{ \frac{240 \ \times \ 10^3}{1.65}} = 381.38$ rad/sec

$\frac{\theta}{\theta_{st}} = \frac{1}{\sqrt{(1-r^2)^2 + (2 \xi r)^2}}$ [NO DAMPER]

$\frac{\theta}{\theta_{st}} = \frac{1}{1 – r_1^2}$ and $\frac{\theta}{\theta_{st}} = \frac{1}{r_2^2 – 1} $

Now: $\theta_{st} = \frac{Mo}{Kt}$

$= \frac{4000}{240 \ \times \ 10^3}$

= 0.0166 rad

$\therefore$ $\frac{\theta}{\theta \ st} = \frac{0.0349}{0.0166} = 2.102$

$2.101 = \frac{1}{1-r_1^2}$

$r_1 \ = \ 0.72$

$2.102 \ = \ \frac{1}{r_2^2 -1}$

$r^2 \ = \ 1.2147$

Now, $r_1 \ = \ \frac{w_1}{w_n}$

$0.72 \times 381.38 \ = \ w_1$

$w_1 \ = \ 274.59$ rad/sec

$r_2 \ = \ \frac{w_2}{w_n}$

$w_2 \ = \ 1.2147 \times 381.38$

$w_2 \ = \ 463.26$ rad/sec

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